October
20, 2006
If
$x>y>0$, show that
$\ln x^2-\ln
y^2<\displaystyle\frac{x^2-y^2}{xy}$
Solution. We can rewrite
the desired inequality as
$\displaystyle\ln\left(\frac{x^2}{y^2}\right)<\frac{x}{y}-\frac{y}{x}$.
Let $f(t)=\ln(t^2)-t+(1/t)$. Then $f'(p)=(2/t)-1-(1/t^2) =
\frac{2t-t^2-1}{t^2}=-\frac{(t-1)^2}{t^2}<0$, for $p>1$. Since
$f(1)=0$, it follows that $f(p)<0$ when $p>1$. This implies the
inequality.